# the base of an isosceles triangle is 8 feet long. if the altitude is 6 feet long and increasing 3 inches per minute, at what rate are the base angles changing?

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1. tanθ = h/4
sec^2θ dθ/dt = 1/4 dh/dt
52 dθ/dt = 1/4 * 3
dθ/dt = 3/4 * 1/52

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oobleck
2. b = base = 8 ft

h = altitude = 6 ft

tan θ = h / ( b / 2 )

tan θ = 2 h / b

θ = arctan ( 2 h / b )

Using the chain rule:

u = 2 h / b

dθ / dh = d [ arctan ( u ) ] / du ∙ du / dh =

1 / ( 1 + u² ) ∙ ( 2 / b ) = 1 / [ 1 + ( 2 h / b )² ] ∙ ( 2 / b ) =

1 / ( 1 + 4 h² / b² ) ∙ ( 2 / b ) = 1 / ( b² / b² + 4 h² / b² ) ∙ ( 2 / b ) =

1 / [ ( b² /+ 4 h² ) / b² ] ∙ ( 2 / b ) = b² / ( b² /+ 4 h² ) ∙ ( 2 / b ) =

b ∙ b / ( b² /+ 4 h² ) ∙ ( 2 / b )

dθ / dh = 2 b / ( b² + 4 h² )

In this case:

dθ / dt = 2 b / ( b² + 4 h² ) = 2 ∙ 8 / ( 8² + 4 ∙ 6² ) =

16 / ( 64 + 4 ∙ 36 ) = 16 / ( 64 + 144 ) = 16 / 208 = 16 ∙ 1 / 16 ∙ 13 = 1 / 13

dθ / dh = 1 / 13 rad / ft

The altitude increasing 3 inches per minute means dh / dt = 3 in / min

1 ft = 12 in

1 in = 1 / 12 ft

3 in = 3 / 12 ft

3 in = 3 ∙ 1 / 3 ∙ 4 ft

3 in = 1 / 4 ft

3 in / min = 1 / 4 ft / min

The altitude increasing dh / dt = 1 / 4 ft / min.

dθ / dt =( dθ / dt) ∙ ( dh / dh )

dθ / dt = ( dθ / dh ) ∙ ( dh / dt )

dθ / dt = ( 1 / 13 ) rad / ft ∙ ( 1 / 4 ) ft / min

dθ / dt = ( 1 / 13 ) rad ∙ ( 1 / 4 ) / min

dθ / dt = 1 / 13 ∙ 4 rad / min

dθ / dt = 1 / 52 rad / min

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